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December 19, 2014

December 19, 2014

Posted by **adam2** on Monday, March 8, 2010 at 10:03pm.

csc (tan^-1 3/3)

- precal -
**Reiny**, Monday, March 8, 2010 at 10:17pmIs this supposed to say,

csc (tan^-1 √3/3) ?

if so, then we have a triangle with opposite as √3 and adjacent as 3, which by Pythagoras gives a hypotenuse of √12

then csc (tan^-1 √3/3) = √12/√3 = √4 = 2

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